By Schur's Lemma above , \(\pi (T)\)  is a representation into \(K=_{T}(V)\) , for \(K\)  a finite field extension of \(k\) .
The representation map \(\pi \colon T\rightarrow (V)\)  is a separable algebraic morphism. Therefore,
it follows by Proposition [1]} that \(\pi ( T)\)  is a \(k\) -analytic multiplicative subgroup of \(K^*\) .
By the Banach Open Mapping Theorem REF , \(\pi \)  is open onto its image so by following diagram
\(\begin{tikzcd}T\times V ["\alpha "] ["\pi "]& \\\pi (T)\times V ["m"] & V\end{tikzcd}\) 
the multiplication by \(\pi (T)\)  is continuous.
The claim now follows directly from Lemma .
